An 8 in. NPS Schedule 80 branch (nozzle) is attached at right angle to a 20-in. NPS Schedule 40 run (header) with a full penetration weld with fillet weld cover. The allowable stress is 13.1 ksi. Design pressure is 600 psi at a design temperature of 900°F. Determine the reinforcing requirements

Question

An 8 in. NPS Schedule 80 branch (nozzle) is attached at right angle to a 20-in. NPS Schedule 40 run (header) with a full penetration weld with fillet weld cover. The allowable stress is 13.1 ksi. Design pressure is 600 psi at a design temperature of 900°F. Determine the reinforcing requirements and pad size, if required.

Accepted Answer

1 . Determine actual and minimum required sizes at the branch run intersection as follows:

[latex]T_{h}=(0.593)(0.875)=0.519 ext { in. }[/latex]

[latex]T_{b}=(0.500)(0.875)=0.438 ext { in. }[/latex]

[latex]\begin{aligned}t_{h} &=P D / 2(S E+P Y)=(600 imes 20) / 2(13,100+0.4 imes 600) \&=0.450 ext { in. }\end{aligned}[/latex]

[latex]\begin{aligned}t_{b} &=P d / 2(S E+P Y)=(600 imes 8.625) / 2(13,100+0.4 imes 600) \&=0.194 ext { in. }\end{aligned}[/latex]

Fillet weld size is the lesser of [latex]0.7 T_{b}[/latex] or [latex]\frac{1}{4}[/latex] in.;

0.7(0.438) = 0.306 in.; use fillet weld size =[latex]\frac{1}{4}[/latex] in.

2 . Determine the horizontal limits of reinforcement from the following, whichever is the greater:

[latex]d_{1}=(8-2 imes 438) / \sin 90^{\circ}=7.125 ext {-in. }[/latex] opening size

[latex]d_{2}=(0.519)+(0.438)+0.5(7.125)=4.519 ext { in. }[/latex] or

[latex]d_{2}=d_{1}=7.125 ext { in. }[/latex]; use [latex]d_{2}=7.125[/latex] in.

3 . Determine the vertical limits of reinforcement from the following, whichever is less:

[latex]L_{4}=2.5(0.519)=1.298 ext { in. }[/latex]

or

[latex]L_{4}=2.5(0.438)+0=1.095 ext { in. }[/latex]; use [latex]L_{4}=1.095[/latex] in.

4 . Determine the reinforcing area required according to Eq. 11.33 as follows:

[latex]A_{1}=t_{h} d_{1}(2-\sin \beta)[/latex] (11.33)

[latex]A_{1}=(0.450)(7.125)\left(2-\sin 90^{\circ} ight)=3.206 ext { in. }{ }^{2}[/latex]

5 . Determine the areas of reinforcement as follows:

Excess in run (header):

[latex]A_{2}=(7.125)(0.519-0.450)=0.492 ext { in. }^{2}[/latex]

Excess in branch (nozzle):

[latex]A_{3}=2(1.095)(0.438-0.194)=0.534 ext { in. }^{2}[/latex]

In fillet welds:

[latex]A_{4}=2\left(\frac{1}{2} ight)\left(\frac{1}{4} ight)^{2}=0.063 ext { in. }^{2}[/latex]

Excess area in [latex]A_{2}+A_{3}+A_{4}=1.089 ext{ in.}^{2}[/latex] This is less than [latex]A_{1}[/latex]; consequently, a pad shall be provided. Determine the thickness of pad based on the pad extending to the horizontal limits of reinforcement.

(7.125) H = 3.206 - 1.089 ; H = 0.297 in.

Assuming [latex]T_{r}=5 / 16[/latex] in. determine minimum pad width.

(2W - 7.125)(0.3125) = 2.117

W = 6.95 in. Use 14 in. × 5/16 in. pad.

[latex]A_{5}=(14-7.125)(0.3125)=2.148 ext { in. }^{2}[/latex]

Available area of reinforcement in [latex]A_{2}+A_{3}+A_{4}+A_{5}=3.237 ext { in. }^{2}[/latex]

This is greater than [latex]A_{1}[/latex] and therefore is satisfactory.

It may be possible to obtain more refinement and a thinner or narrower plate by reevaluating the vertical limits by setting [latex]T_{r}=0.312[/latex] in. and by including the outer fillet welds if they lie within the horizontal reinforcement limits. This recalculation may reduce the pad thickness and/or the pad width. However, a 14 -in. by 5/16-in. pad is satisfactory.

Question 11.12: An 8 in. NPS Schedule 80 branch (nozzle) is attached at righ...

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